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Bounded modern-physics survey · Interactive lesson

Fusion, the Coulomb barrier, and tunnelling

The Sun's protons meet each other with about 2 keV against a 600 keV wall, and fuse anyway. This topic is the calculation that explains how: a barrier set by charge and nuclear size, a tunnelling exponent from a WKB integral, and a Maxwell tail, whose product decides where stars and tokamaks actually burn.

UPIII-11.08 · Lesson 8 of 9 in this unit. Position is not a completion score.
Starting model: Fusion, the Coulomb barrier, and tunnellingLog-log plot against centre-of-mass energy from 1 keV to 1 MeV; the vertical axis runs from 1 at the top to e⁻⁴⁰. Dashed: Maxwell factor e^(−E/kT). Faint: tunnelling factor exp(−√(E_G/E)), stopped at the shaded hydrogen-isotope barriers. Solid: their product, peaking (dot) at E₀ = 6.1 keV, where it is e^(−13.5).Peak e(−13.5) vs over 0.5 MeV e(−370)Maxwell tailtunnellingbarrier3kT/21 keV10 keV100 keV1 MeVe⁻⁴⁰
Start with a picture. Change the model in the lab below.

Bounded modern-physics survey. This modern-physics material is an introductory survey, not a complete course in relativity, quantum, atomic, nuclear or particle physics. Your university determines the required depth and assessment.

1

Build the model

Stars fuse at a hundredth of the barrier height, in the thermal tail, by tunnelling.

The key idea

Fusion is the merging of two light nuclei into a more tightly bound one; at thermal energies it happens only when a pair from the high-energy tail of the Maxwell distribution tunnels through the Coulomb barrier that keeps it out of range of the strong force.

Example. Two protons in the solar core (kT = 1.35 keV) face a 0.60 MeV barrier. The pairs that fuse cluster near 6.1 keV, where tunnelling costs e(−9.0) ≈ 1.2 × 10⁻⁴ and the Maxwell tail e(−4.5), instead of the e(−444) of climbing over the top.

The complete physical model and assumptions

Fusion is where the energy ledger says yes and classical mechanics says no. Light nuclei release MeV when they merge, 17.6 MeV for deuterium and tritium, because binding energy per nucleon climbs toward ⁴He. To merge, though, they must reach the femtometre range of the strong force, and outside it they repel with the full Coulomb force, so the potential peaks at contact at a few hundred keV: 0.44 MeV for D-T, 0.60 MeV for two protons. The solar core gives a pair about 2 keV, and the classical Maxwell factor for climbing the barrier is e(−444): not one proton in the Sun would ever fuse. Two exponentials close the gap. Tunnelling passes a pair of centre-of-mass energy E with probability exp(−√(EG/E)), where the Gamow energy EG = 2mᵣ c²(παZ₁Z₂)² folds charge and mass into one number, while the thermal distribution weights each energy E by the Boltzmann factor e(−E/kT). One factor rises and the other falls, so their product is a narrow Gamow peak at E₀ = (EG(kT)²/4)¹⁄³: about 6 keV in the Sun, several times kT yet a hundredth of the barrier. That predicts where reactions happen and why their rate climbs so steeply with temperature. It stops at the nuclear factor S(E): it cannot say why p-p must wait for a weak decay that leaves its S factor some 10²⁵ times below D-T's, nor place the resonance that makes D-T the fuel of choice.

1Two reactions, one climb up the binding curve

Fusing light nuclei releases energy because binding energy per nucleon climbs steeply, from 1.11 MeV in ²H to 7.07 MeV in ⁴He. The reaction behind every tokamak design is ²H + ³H → ⁴He + n. With atomic masses the two electrons of the reactant atoms match the two in helium, so Q = (2.014102 + 3.016049 − 4.002603 − 1.008665) u × 931.494 MeV u⁻¹ = 17.6 MeV. The products leave with equal and opposite momenta, so the kinetic energy divides inversely with mass: 14.1 MeV to the neutron, 3.5 MeV to the alpha particle. The Sun starts instead with p + p → ²H + e⁺ + νₑ, worth only 0.42 MeV, in which one proton must become a neutron by the weak interaction during the encounter. Its chain ends at 4 ¹H → ⁴He + 2e⁺ + 2νₑ, 26.7 MeV per helium nucleus once the positrons annihilate. Every one of these is exothermic by MeV. Energy is not what holds fusion back.

2The barrier peaks where the strong force takes over

Outside contact, two nuclei feel only their Coulomb repulsion, V(r) = Z₁Z₂(1.440 MeV fm)/r. The strong force reaches a femtometre or two, so it switches on only when the surfaces touch, at r ≈ R₁ + R₂ with R = 1.2A¹⁄³ fm, and the potential then drops into a well tens of MeV deep. The barrier therefore peaks at contact: VC = 1.440/3.24 = 0.44 MeV for D-T and 1.440/2.40 = 0.60 MeV for p-p, but 2.5 MeV for p + ¹⁴N, since the height grows as Z₁Z₂ and falls only as A¹⁄³. The energy that counts is the centre-of-mass kinetic energy E: the motion of the centre of mass is conserved and cannot be spent against the barrier. Classically a pair turns back at rc = Z₁Z₂(1.440 MeV fm)/E. Two protons at 6.1 keV stop 237 fm apart, a hundred times the contact distance, and that 237 fm of forbidden region is what any quantum rescue has to cross.

3Thermal energies fall hundreds of times short

At the centre of the Sun, T = 1.57 × 10⁷ K, so kT = 1.35 keV and the mean relative kinetic energy (3/2)kT is 2.0 keV, 300 times below the p-p barrier. A D-T plasma at 1.5 × 10⁸ K, extreme by any terrestrial standard, has kT = 12.9 keV, still 34 times below 0.44 MeV. Pairs are not all average. Relative energies follow the normalised Maxwell–Boltzmann distribution f(E) = (2/√π)(kT)⁻³⁄² √E e(−E/kT), with ∫₀^∞ f dE = 1, and for VC ≫ kT the fraction of pairs above the barrier is about (2/√π)√(VC/kT) e(−VC/kT). For the Sun, VC/kT = 444 and that fraction is 2 × 10⁻¹⁹². Even if all of the Sun's 10⁵⁷ protons sat at core temperature, the expected number above the barrier at any instant would be about 2 × 10⁻¹³⁵. Raising the mean energy to the barrier instead would take 4.6 × 10⁹ K. On classical physics the Sun cannot shine.

4Tunnelling: the WKB integral gives the Gamow factor

Inside the forbidden region the relative wavefunction decays at the local rate κ(r) = √(2mᵣ(V(r) − E))/ℏ, with mᵣ = m₁m₂/(m₁ + m₂) the reduced mass, and the transmission probability is P ≈ exp(−2∫κ dr), taken from contact out to rc. With V = Erc/r, and contact far inside the turning point, the lower limit can be set to zero: the integral of √(rc/r − 1) from 0 to rc is πrc/2. The exponent becomes πrc√(2mᵣE)/ℏ, which rearranges to √(EG/E) with EG = 2mᵣ c²(παZ₁Z₂)². For p-p, EG = 493 keV, so at 6.1 keV P ≈ e(−9.0) = 1.2 × 10⁻⁴: small, not zero. A size check agrees. At that energy ℏ/p = 83 fm, and the barrier is 237 fm thick, under three reduced wavelengths. Correspondence holds as well: EG ∝ 1/ℏ², so as ℏ → 0 the exponent diverges for every E below the barrier and the classical verdict returns.

5Maxwell tail times tunnelling: the Gamow peak

The rate per pair is σv averaged over the thermal distribution. Factor the cross section as σ(E) = (S(E)/E) exp(−√(EG/E)): 1/E carries the geometric area πλ̄², the exponential carries the barrier, and S(E) carries the nuclear physics, slowly varying for a non-resonant reaction. Then ⟨σv⟩ ∝ ∫ S(E) exp(−E/kT − √(EG/E)) dE, one exponential falling with E and the other rising. Setting the derivative of the exponent to zero, 1/kT = ½EG¹⁄²E⁻³⁄², gives E₀ = (EG(kT)²/4)¹⁄³, and at E₀ the exponent equals exactly 3E₀/kT. For solar p-p, E₀ = 6.1 keV: 4.5 kT and 1% of the barrier, inside a window of 1/e full width 4√(E₀kT/3) = 6.6 keV. The suppression is e(−13.5) = 1.4 × 10⁻⁶ instead of e(−444), a gain of 10¹⁸⁷. Because 3E₀/kT falls as T⁻¹⁄³, and the prefactor goes as T⁻²⁄³, the rate climbs locally as Tν with ν = (3E₀/kT − 2)/3 = 3.8.

6What the model fixes, and what it leaves to nuclear physics

The barrier and the tail fix where reactions happen and how steeply the rate climbs with temperature. They do not fix how many: that is S(E), and this survey cannot predict it. For D-T, S is about 10⁴ keV b, and a ⁵He resonance lifts the cross section to about 5 b near 64 keV, a feature no smooth S can produce. For p-p, S is about 4 × 10⁻²² keV b, some 10²⁵ times smaller, because a proton must turn into a neutron by the weak interaction while the pair overlaps. That is why a proton in the solar core waits about 10¹⁰ years to fuse, and why the Sun lasts. Smaller effects sharpen the estimate without overturning it: finite nuclear size shortens the tunnel, cutting the p-p exponent by the fraction (4/π)√((R₁ + R₂)/rc) = 13% at 6.1 keV, and electron screening raises solar rates by a few per cent. Burning is a further question again: a reactor needs enough density and confinement time for fusion power to beat its losses, the Lawson condition.

2

Key equations

Read the units, assumptions and limits before substituting.

Coulomb barrier at contact

VC = Z₁Z₂e²/[4πε₀(R₁ + R₂)]

The height a pair must climb classically: 0.44 MeV for D-T, 0.60 MeV for p-p, 2.5 MeV for p + ¹⁴N. It grows with charge far faster than it falls with size.

Z₁, Z₂ charge numbers; R = r₀A¹⁄³ with r₀ ≈ 1.2 fm; e²/(4πε₀) = 1.440 MeV fm; VC in MeV (J in SI)

Gamow energy

EG = 2mᵣ c²(παZ₁Z₂)²

One number for the whole barrier: 493 keV for p-p, 1182 keV for D-T, 45 MeV for p + ¹⁴N. Growing as Z₁²Z₂² is why each heavier fuel needs a much hotter core.

mᵣ = m₁m₂/(m₁ + m₂) the reduced mass; α = e²/(4πε₀ℏc) ≈ 1/137; EG in keV (J in SI)

Tunnelling probability below the barrier

P(E) ≈ exp(−√(EG/E))

Rises steeply with energy: for p-p it is e(−19.1) at E = 1.35 keV but e(−9.0) at 6.1 keV, a factor of 2 × 10⁴ for less than five times the energy.

E the centre-of-mass kinetic energy, same unit as EG; WKB through a point-charge Coulomb wall, valid for E ≪ VC

Cross section, factored

σ(E) = (S(E)/E) exp(−√(EG/E))

Separates geometry (1/E, from πλ̄²), barrier and nuclear physics. S alone carries the 10²⁵ gap between D-T and p-p that the barrier cannot explain.

σ in m² (1 b = 10⁻²⁸ m²); S(E) the astrophysical S factor in keV b, slowly varying away from resonances

Thermal reaction rate per pair

⟨σv⟩ = √(8/(πmᵣ)) (kT)⁻³⁄² ∫₀^∞ S(E) exp(−E/kT − √(EG/E)) dE

Reactions per unit volume per second are n₁n₂⟨σv⟩, halved for identical nuclei. The integrand is the Maxwell tail times the tunnelling factor: the source of the Gamow peak.

⟨σv⟩ in m³ s⁻¹ with E, kT in J, mᵣ in kg and S in J m²; from the normalised Maxwell–Boltzmann distribution

Gamow peak

E₀ = (EG(kT)²/4)¹⁄³

Solar p-p at kT = 1.35 keV: E₀ = 6.1 keV, or 4.5 kT. The integrand there is e(−3E₀/kT) = e(−13.5), against e(−444) for going over the top.

E₀, EG and kT in one energy unit; assumes S(E) nearly constant across the peak and E₀ ≪ VC
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

1.35 keV
1 to 20 keV; steps of 0.05.
493 keV
400 to 1300 keV; steps of 1.

Try this. Start at solar p-p, kT = 1.35 keV and EG = 493 keV. Raise EG to 1182 keV for D-T and the peak slides right and sinks to e(−18). Then raise kT to 10 keV, a tokamak: the peak climbs to 31 keV and e(−9.3), still far left of the barrier band.

Live results

GAMOW PEAK E₀
6.1 keV
PEAK WIDTH Δ
6.6 keV
PEAK EXPONENT τ = 3E₀/kT
13.5
TEMPERATURE POWER ν
3.8

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics III modelLog-log plot against centre-of-mass energy from 1 keV to 1 MeV; the vertical axis runs from 1 at the top to e⁻⁴⁰. Dashed: Maxwell factor e^(−E/kT). Faint: tunnelling factor exp(−√(E_G/E)), stopped at the shaded hydrogen-isotope barriers. Solid: their product, peaking (dot) at E₀ = 6.1 keV, where it is e^(−13.5).Peak e(−13.5) vs over 0.5 MeV e(−370)Maxwell tailtunnellingbarrier3kT/21 keV10 keV100 keV1 MeVe⁻⁴⁰
Log-log plot against centre-of-mass energy from 1 keV to 1 MeV; the vertical axis runs from 1 at the top to e⁻⁴⁰. Dashed: Maxwell factor e(−E/kT). Faint: tunnelling factor exp(−√(EG/E)), stopped at the shaded hydrogen-isotope barriers. Solid: their product, peaking (dot) at E₀ = 6.1 keV, where it is e(−13.5).

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Thermal energy kT: 1.35 keV; Gamow energy EG: 493 keV

Baseline 1.35; choose 1 to 20, in steps of 0.05.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The tempting estimate sets the mean thermal energy equal to the barrier, (3/2)kT = VC, and concludes that p-p fusion needs 4.6 × 10⁹ K, three hundred times the Sun's core temperature, and that a D-T reactor needs 3.4 × 10⁹ K. It is wrong on two counts. The rate is an integral over the whole Maxwell distribution, not a property of the average pair, so rare pairs far out in the tail can dominate it. And a pair never has to reach the top: it tunnels, with a probability that rises steeply with E. Reactions come from the product of the two, at E₀ ≈ 4.5 kT in the Sun, where the combined suppression is e(−13.5) rather than e(−444). The opposite shortcut fails as well: evaluating the integrand at the mean energy gives e(−17.1), about 36 times below the peak, so the Maxwell tail is not optional either.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

A beam of deuterons with lab kinetic energy 100 keV strikes a tritium target at rest. Using R = 1.2A¹⁄³ fm and e²/(4πε₀) = 1.440 MeV fm, find the D-T Coulomb barrier, the centre-of-mass energy of a collision, and the classical distance of closest approach. Could the pair reach contact classically?

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

MediumExample 2

A D-T plasma is held at kT = 10.0 keV. With EG = 1182 keV for the pair and a barrier of 0.444 MeV, find the Gamow-peak energy, the tunnelling probability and Maxwell factor there, and the width of the window. Compare the peak integrand with the Maxwell factor for climbing over the barrier.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

A star slightly more massive than the Sun has a core at 2.0 × 10⁷ K. Compare p + p (EG = 493 keV) with the CNO-cycle bottleneck p + ¹⁴N (reduced mass 875 MeV/c²). For each, find EG where needed, the Gamow peak, the peak exponent τ = 3E₀/kT and the local temperature power ν in ⟨σv⟩ ∝ Tν. Then find the factor by which each rate rises if the core warms by 5%.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

In the solar core kT = 1.35 keV. For two protons the Gamow energy is EG = 493 keV and the Coulomb barrier is about 0.60 MeV. Near what centre-of-mass energy do the collisions that actually fuse take place?
7

Review & continue

Fusion is the merging of two light nuclei into a more tightly bound one; at thermal energies it happens only when a pair from the high-energy tail of the Maxwell distribution tunnels through the Coulomb barrier that keeps it out of range of the strong force.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Optional supporting lessons

These chapter resources can help with earlier ideas; they are not additional required Physics III completion tasks.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.